| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Tangent with specified gradient |
| Difficulty | Moderate -0.3 This is a straightforward tangent problem requiring students to equate the line to the curve, use the discriminant condition (b²-4ac=0) for tangency, then find the point of contact. While it involves multiple steps (differentiation, substitution, solving quadratic), these are standard techniques with no conceptual difficulty beyond typical A-level expectations. |
| Spec | 1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{x^2}{4} = \frac{x}{k} + k \rightarrow kx^2 - 4x - 4k^2 = 0\) | M1 | Eliminates \(x\) or \(y\) completely |
| Uses \(b^2 - 4ac \rightarrow k = -1\) | M1 | Uses \(b^2 - 4ac\) for a quadratic \(= 0\) |
| A1 | co, nb \(a,b,c\) must not be \(f(x)\) | |
| [3] | ||
| (calculus \(\frac{1}{k} = \frac{2x}{4}\)) | B1 | |
| \(\rightarrow x = \frac{2}{k}\), \(y = \frac{1}{k^2}\) M1 \(\rightarrow k = -1\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = -x - 1\), \(4y = x^2\) | M1 | Elimination of \(x\) or \(y\) |
| \(\rightarrow x^2 + 4x + 4 = 0\) | M1 A1 | Solution of equation, co |
| \(\rightarrow P(-2, 1)\) | ||
| [3] |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{x^2}{4} = \frac{x}{k} + k \rightarrow kx^2 - 4x - 4k^2 = 0$ | M1 | Eliminates $x$ or $y$ completely |
| Uses $b^2 - 4ac \rightarrow k = -1$ | M1 | Uses $b^2 - 4ac$ for a quadratic $= 0$ |
| | A1 | co, nb $a,b,c$ must not be $f(x)$ |
| **[3]** | | |
| (calculus $\frac{1}{k} = \frac{2x}{4}$) | B1 | |
| $\rightarrow x = \frac{2}{k}$, $y = \frac{1}{k^2}$ M1 $\rightarrow k = -1$ | A1 | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = -x - 1$, $4y = x^2$ | M1 | Elimination of $x$ or $y$ |
| $\rightarrow x^2 + 4x + 4 = 0$ | M1 A1 | Solution of equation, co |
| $\rightarrow P(-2, 1)$ | | |
| **[3]** | | |
---4 The line $y = \frac { x } { k } + k$, where $k$ is a constant, is a tangent to the curve $4 y = x ^ { 2 }$ at the point $P$. Find\\
(i) the value of $k$,\\
(ii) the coordinates of $P$.
\hfill \mbox{\textit{CAIE P1 2012 Q4 [6]}}